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Kisachek [45]
3 years ago
8

|4| = A. 4 B. 1 C. –4 D. 0

Mathematics
2 answers:
kvv77 [185]3 years ago
5 0

I believe the correct answer is A. This is because it's asking for the absolute answer of 4, and 4=4. I hope this helps!

zhuklara [117]3 years ago
4 0

the answer to this problem is 4

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A tank with a capacity of 1600 L is full of a mixture of water and chlorine with a concentration of 0.0125 g of chlorine per lit
Veronika [31]

Answer:

y(t) = 20 [1600^(-5/3)] x (1600-24t)^ (5/3)

Step-by-step explanation:

1) Identify the problem

This is a differential equation problem

On this case the amount of liquid in the tank at time t is 1600−24t. (When the process begin, t=0 ) The reason of this is because the liquid is entering at 16 litres per second and leaving at 40 litres per second.

2) Define notation

y = amount of chlorine in the tank at time t,

Based on this definition, the concentration of chlorine at time t is y/(1600−24t) g/ L.

Since liquid is leaving the tank at 40L/s, the rate at which chlorine is leaving at time t is 40y/(1600−24t) (g/s).

For this we can find the differential equation

dy/dt = - (40 y)/ (1600 -24 t)

The equation above is a separable Differential equation. For this case the initial condition is y(0)=(1600L )(0.0125 gr/L) = 20 gr

3) Solve the differential equation

We can rewrite the differential equation like this:

dy/40y = -  (dt)/ (1600-24t)

And integrating on both sides we have:

(1/40) ln |y| = (1/24) ln (|1600-24t|) + C

Multiplying both sides by 40

ln |y| = (40/24) ln (|1600 -24t|) + C

And simplifying

ln |y| = (5/3) ln (|1600 -24t|) + C

Then exponentiating both sides:

e^ [ln |y|]= e^ [(5/3) ln (|1600-24t|) + C]

with e^c = C , we have this:

y(t) = C (1600-24t)^ (5/3)

4) Use the initial condition to find C

Since y(0) = 20 gr

20 = C (1600 -24x0)^ (5/3)

Solving for C we got

C = 20 / [1600^(5/3)] =  20 [1600^(-5/3)]

Finally the amount of chlorine in the tank as a function of time, would be given by this formula:

y(t) = 20 [1600^(-5/3)] x (1600-24t)^ (5/3)

7 0
3 years ago
Se corta una manzana en 12 partes, se comen ocho partes. Qué fracción simplificada a su mínima expresión representa la cantidad
lakkis [162]

Answer:

A. 1/3

Step-by-step explanation:

Dado que:

una manzana se corta en 12 partes y se comen ocho partes.

La fracción de manzanas consumidas = 8/12

= 2/3

La fracción de manzanas que quedan = valor original de la manzana que se corta - la fracción de la manzana que se come

La fracción de manzanas que quedan  =1-  \dfrac{8}{12}

La fracción de manzanas que quedan  =\dfrac{12-8}{12}

La fracción de manzanas que quedan =\dfrac{4}{12}

A la fracción más baja; obtenemos = 1/3

6 0
3 years ago
Heres another 56 points for free. Good Luck :)
murzikaleks [220]

Answer:

Thank you

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
What is the value of x in 2(5x) = 14?
Ray Of Light [21]

Answer:

x= 7/5

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
50,000 x (2 + 5/100)<br> 50,000 x (3 + 5/100)<br> 50,000 x (4 + 5/100)<br> 50,000 x (5 + 5/100)
ELEN [110]

Answer:

Solution for What is 5 percent of 50000:

5 percent *50000 =

( 5:100)*50000 =

( 5*50000):100 =

250000:100 = 2500

Step-by-step explanation:

7 0
3 years ago
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